new test - october 20, 2014 - david delgado...of the population take less than 5 minutes to get to...

New test - October 20, 2014 [642 marks]

1a. [3 marks]

Let and .

Find AB .

Markschemeevidence of multiplying (M1)

e.g. one correct element,

A2 N3

Note: Award A1 for three correct elements.

[3 marks]

Examiners reportThe large majority of candidates answered this question successfully. There were only a small number of candidates who seemed tohave never worked with matrices before. Occasionally a candidate would incorrectly approach part (b) by finding an inverse ofmatrix A.

1b. [3 marks]Given that , find X.

Markschemefinding (A1)

adding 2 to both sides (may be seen first) (M1)

e.g.

A1 N2

[3 marks]

Examiners reportThe large majority of candidates answered this question successfully. There were only a small number of candidates who seemed tohave never worked with matrices before. Occasionally a candidate would incorrectly approach part (b) by finding an inverse ofmatrix A.

2a. [3 marks]

The following table shows the probability distribution of a discrete random variable X .

Find the value of k .

Markschemeevidence of summing to 1 (M1)

e.g.

correct working (A1)

e.g.

A1 N2

[3 marks]

Examiners reportOverall, this question was very well done. A few candidates left this question blank, or used methods which would indicate they wereunfamiliar with discrete random variables. In part (b), there were a good number of candidates who set up their work correctly, butthen had trouble adding or multiplying decimals without a calculator. A common type of error for these candidates was .

2b. [3 marks]Find .

Markschemecorrect substitution into formula (A1)

e.g.

correct working

e.g. (A1)

= 3.3 A1 N2

[3 marks]

Examiners reportOverall, this question was very well done. A few candidates left this question blank, or used methods which would indicate they wereunfamiliar with discrete random variables. In part (b), there were a good number of candidates who set up their work correctly, butthen had trouble adding or multiplying decimals without a calculator. A common type of error for these candidates was .

3a. [4 marks]

The weekly wages (in dollars) of 80 employees are displayed in the cumulative frequency curve below.

(i) Write down the median weekly wage.

(ii) Find the interquartile range of the weekly wages.

Markscheme(i) median weekly wage (dollars) A1 N1

(ii) lower quartile , upper quartile (A1)(A1)

(dollars) (accept any notation suggesting interval to ) A1 N3

Note: Exception to the FT rule. Award A1(FT) for an incorrect IQR only if both quartiles are explicitly noted.

[4 marks]

Examiners reportMany candidates answered this question completely correctly, earning full marks in all parts of the question. In parts (a) and (b), therewere some who gave the frequency values on the y-axis, rather than the wages on the x-axis, as their quartiles and interquartile range.

3b. [3 marks]The box-and-whisker plot below displays the weekly wages of the employees.

Write down the value of

(i) ;

(ii) ;

(iii) .

Markscheme(i) (dollars) A1 N1

(ii) (dollars) A1 N1

(iii) (dollars) A1 N1

[3 marks]

Examiners reportMany candidates answered this question completely correctly, earning full marks in all parts of the question. In parts (a) and (b), therewere some who gave the frequency values on the y-axis, rather than the wages on the x-axis, as their quartiles and interquartile range.

3c. [3 marks]Employees are paid per hour.

Find the median number of hours worked per week.

Markschemevalid approach (M1)

e.g.

correct substitution (A1)

e.g.

median hours per week A1 N2

[3 marks]

Examiners reportFor part (c), the majority of candidates seemed to understand what was required, though there were a few who used an extreme valuesuch as , rather than the median value.

3d. [5 marks]Employees are paid per hour.

Find the number of employees who work more than hours per week.

Markschemeattempt to find wages for 25 hours per week (M1)

e.g.


e.g.

finding wages (A1)

65 people (earn 500 ) (A1)

15 people (work more than 25 hours) A1 N3

[5 marks]

Examiners reportIn part (d), some candidates simply answered , which is the number of workers earning or less, rather than finding the numberof workers who earned more than . It was interesting to note that quite a few candidates gave their final answer as , rather than

.

4. [8 marks]In a large city, the time taken to travel to work is normally distributed with mean and standard deviation . It is found that of the population take less than 5 minutes to get to work, and take less than 25 minutes.

Find the value of and of .

Markschemecorrect z-values (A1)(A1)

,

attempt to set up their equations, must involve z-values, not % (M1)

e.g. one correct equation

two correct equations A1A1

e.g. ,

attempt to solve their equations (M1)

e.g. substitution, matrices, one correct value

,

, A1A1 N4

[8 marks]

Examiners reportA standard question for which well-prepared candidates frequently earned all eight marks. Common errors included the use ofpercentages rather than z-values and the inability to find the negative z-value. Others had correct equations but were not able to usetheir GDC to solve them and ultimately made errors in their algebra.

5a. [2 marks]At a large school, students are required to learn at least one language, Spanish or French. It is known that of the studentslearn Spanish, and learn French.

Find the percentage of students who learn both Spanish and French.


e.g. Venn diagram with intersection, union formula,

(accept ) A1 N2

[2 marks]

Examiners reportParts (a) and (b) were generally done well although some candidates left answers as decimals rather than the required percentages.

5b. [2 marks]At a large school, students are required to learn at least one language, Spanish or French. It is known that of the studentslearn Spanish, and learn French.

Find the percentage of students who learn Spanish, but not French.

Markschemevalid approach involving subtraction (M1)

e.g. Venn diagram,

60 (accept ) A1 N2

[2 marks]

Examiners reportParts (a) and (b) were generally done well although some candidates left answers as decimals rather than the required percentages.

5c. [5 marks]At a large school, students are required to learn at least one language, Spanish or French. It is known that of the studentslearn Spanish, and learn French.

At this school, of the students are girls, and of the girls learn Spanish.

A student is chosen at random. Let G be the event that the student is a girl, and let S be the event that the student learns Spanish.

(i) Find .

(ii) Show that G and S are not independent.

Markscheme(i) valid approach (M1)

e.g. tree diagram, multiplying probabilities,

correct calculation (A1)

e.g.

(exact) A1 N3

(ii) valid reasoning, with words, symbols or numbers (seen anywhere) R1

e.g. , , not equal,

one correct value A1

e.g. , ,

G and S are not independent AG N0

[5 marks]

Examiners reportIn part (c) (i), candidates failed to find the intersection of the events as, in general, they multiplied probabilities, assuming the eventswere independent or they incorrectly attempted to use the union formula. Independence in (c) (ii) caused difficulty with somecandidates attempting to use the conditions for mutually exclusive events while others assumed the events were independent in part (i)and then found by multiplying .

5d. [6 marks]At a large school, students are required to learn at least one language, Spanish or French. It is known that of the studentslearn Spanish, and learn French.

At this school, of the students are girls, and of the girls learn Spanish.

A boy is chosen at random. Find the probability that he learns Spanish.

MarkschemeMETHOD 1

are boys (seen anywhere) A1

e.g.

appropriate approach (M1)

e.g.

correct approach to find P(boy and Spanish) (A1)

e.g. , , 0.308


e.g. ,

correct manipulation (A1)

e.g.

,

A1 N3

[6 marks]

METHOD 2

are boys (seen anywhere) A1

e.g. 0.48 used in tree diagram


e.g. tree diagram

correctly labelled branches on tree diagram (A1)

e.g. first branches are boy/girl, second branches are Spanish/not Spanish


e.g.

correct manipulation (A1)

e.g. ,

,

[6 marks]

Examiners reportPart (d) proved quite challenging as a great majority could only find the probability of being a boy. Those who did attempt it, andsuccessfully connected the problem with conditional probability, often had difficulties in reaching the correct final answer.

6a. [5 marks]

Jar A contains three red marbles and five green marbles. Two marbles are drawn from the jar, one after the other, without replacement.

Find the probability that

(i) none of the marbles are green;

(ii) exactly one marble is green.

Markscheme(i) attempt to find (M1)

eg , ,

A1 N2

(ii) attempt to find (M1)

eg , ,

recognizing two ways to get one red, one green (M1)

eg , ,

A1 N2

[5 marks]

Examiners reportMany candidates correctly found the probability of selecting no green marbles in two draws, although some candidates treated thesecond draw as if replacing the first. When finding the probability for exactly one green marble, candidates often failed to recognizetwo pathways for selecting one of each color.

6b. [3 marks]Find the expected number of green marbles drawn from the jar.

Markscheme (seen anywhere) (A1)

correct substitution into formula for A1

eg ,

expected number of green marbles is A1 N2

[3 marks]

Examiners reportFew candidates understood the concept of expected value in this context, often leaving this blank or treating as if a binomialexperiment. Successful candidates often made a distribution table before making the final calculation.

6c. [2 marks]

Jar B contains six red marbles and two green marbles. A fair six-sided die is tossed. If the score is or , a marble is drawn from jar A.Otherwise, a marble is drawn from jar B.

(i) Write down the probability that the marble is drawn from jar B.

(ii) Given that the marble was drawn from jar B, write down the probability that it is red.

Markscheme(i) A1 N1

(ii) A1 N1

[2 marks]

Examiners reportMost candidates answered part (c) correctly. However, many overcomplicated (c)(ii) by using the conditional probability formula.Those with a clear understanding of the concept easily followed the “write down” instruction.

6d. [6 marks]Given that the marble is red, find the probability that it was drawn from jar A.

Markschemerecognizing conditional probability (M1)

eg , , tree diagram

attempt to multiply along either branch (may be seen on diagram) (M1)

eg

attempt to multiply along other branch (M1)

eg

adding the probabilities of two mutually exclusive paths (A1)

eg

correct substitution

eg , A1

A1 N3

[6 marks]

Examiners reportOnly a handful of candidates correctly applied conditional probability to find in part (d). While some wrote down theformula, or drew a tree diagram, few correctly calculated . A common error was to combine the marbles in the two jars togive .

7a. [2 marks]

Consider the following cumulative frequency table.

Find the value of .


eg ,

A1 N2

[2 marks]

Examiners reportCandidates had little problem determining a missing frequency from a cumulative frequency table.

7b. [4 marks]Find

(i) the mean;

(ii) the variance.

Markscheme(i) mean A2 N2

(ii) recognizing that variance is (sd) (M1)

eg , ,

A1 N2

[4 marks]

Examiners reportIn part (b), few used the GDC to their advantage to correctly find the mean and variance. There were numerous unsuccessful attemptsat using the formulae for mean and variance, most resulting in algebraic errors along the way. Candidates recognized the concept ofvariance but were often unable to determine what value should be squared.

2

8. [7 marks]A random variable is normally distributed with and .

Find the interquartile range of .

Markschemerecognizing one quartile probability (may be seen in a sketch) (M1)

eg ,

finding standardized value for either quartile (A1)

eg ,

attempt to set up equation (must be with -values) (M1)

eg ,

one correct quartile

eg ,


eg other correct quartile,

valid approach for IQR (seen anywhere) (A1)

eg ,

IQR A1 N4

[7 marks]

Examiners reportThis was an accessible problem that created difficulties for candidates. Although they recognized and often wrote down a formula forIQR, most did not understand the conceptual nature of the first and third quartiles. Those who did could solve the problem effectivelyusing their GDC in relatively few steps. Candidates that were able to start this question often drew the normal curve and gave quartilevalues at and . This generally led to a solution which while wrong, was also clearly inadequate for the indicated 7 marks.

9. [5 marks]

A Ferris wheel with diameter metres rotates clockwise at a constant speed. The wheel completes rotations every hour. The bottom ofthe wheel is metres above the ground.

A seat starts at the bottom of the wheel.

After t minutes, the height metres above the ground of the seat is given by

In one rotation of the wheel, find the probability that a randomly selected seat is at least metres above the ground.

Markschemesetting up inequality (accept equation) (M1)

eg , , sketch of graph with line

any two correct values for t (seen anywhere) A1A1

eg , , ,

valid approach M1

eg , , ,

A1 N2

[5 marks]

Examiners reportPart (e) was very poorly done for those who attempted the question and most did not make the connection between height, time andprobability. The idea of linking probability with a real-life scenario proved beyond most candidates. That said, there were a few novelapproaches from the strongest of candidates using circles and angles to solve this part of question 10.

10a. [4 marks]

A running club organizes a race to select girls to represent the club in a competition.

The times taken by the group of girls to complete the race are shown in the table below.

Find the value of and of .

Markschemeattempt to find (M1)

eg ,

A1 N2

attempt to find (M1)

eg ,

A1 N2

[4 marks]

Examiners reportOverall, candidates were very successful in parts (a), (b) and (c) of this question. Most of the errors in these parts had to do withcandidates not understanding terms such as "at least" or "less than".

10b. [3 marks]A girl is chosen at random.

(i) Find the probability that the time she takes is less than minutes.

(ii) Find the probability that the time she takes is at least minutes.

Markscheme(i) A1 N1

(ii) valid approach (M1)

eg ,

A1 N2

[3 marks]


10c. [4 marks]A girl is selected for the competition if she takes less than minutes to complete the race.

Given that of the girls are not selected,

(i) find the number of girls who are not selected;

(ii) find .

Markscheme(i) attempt to find number of girls (M1)

eg ,

are not selected A1 N2

(ii) are selected (A1)

A1 N2

[4 marks]


10d. [4 marks]Girls who are not selected, but took less than minutes to complete the race, are allowed another chance to be selected. Thenew times taken by these girls are shown in the cumulative frequency diagram below.

(i) Write down the number of girls who were allowed another chance.

(ii) Find the percentage of the whole group who were selected.

Markscheme(i) given second chance A1 N1

(ii) took less than minutes (A1)

attempt to find their selected total (may be seen in calculation) (M1)

eg , their answer from (i)

( ) A1 N3

[4 marks]

Examiners reportPart (d) was quite challenging for candidates, who may not have read the question carefully and studied the values in the diagram.Many seemed confused by the idea that not all the girls who were given a second chance were selected. In part (d)(ii), many did notfind the percentage of the whole group, but rather the percentage of the girls who were given a second chance.

11a. [3 marks]

The random variable is normally distributed with mean and standard deviation .

Find .

Markschemeevidence of appropriate approach (M1)

eg

(A1)

A1 N3

[3 marks]

Examiners reportThe normal distribution was handled better than in previous years with many candidates successful in both parts and very few blankresponses. Some candidates used tables and -scores while others used the GDC directly; the GDC approach earned full marks moreoften than the -score approach.

11b. [3 marks]Given that , find the value of .

Markscheme-score for is (A1)

valid approach (must be with -values) (M1)

eg using inverse normal,

A1 N3

[3 marks]

Examiners reportThe normal distribution was handled better than in previous years with many candidates successful in both parts and very few blankresponses. Some candidates used tables and -scores while others used the GDC directly; the GDC approach earned full marks moreoften than the -score approach. A common error in part (b) was to set the expression for -score equal to the probability. Manycandidates had difficulty giving answers correct to three significant figures; this was particularly an issue if no working was shown.

12a. [5 marks]

A bag contains four gold balls and six silver balls.

Two balls are drawn at random from the bag, with replacement. Let be the number of gold balls drawn from the bag.

(i) Find .

(ii) Find .

MarkschemeMETHOD 1

(i) appropriate approach (M1)

eg , ,

A1 N2

(ii) multiplying one pair of gold and silver probabilities (M1)

eg , , 0.24

adding the product of both pairs of gold and silver probabilities (M1)

eg ,

A1 N3

METHOD 2

(i) evidence of recognizing binomial (may be seen in part (ii)) (M1)

eg ,

correct probability for use in binomial (A1)

eg , ,

A1 N3

(ii) correct set up (A1)

eg

A1 N2

[5 marks]

Examiners reportParts (a)(i) and (ii) were generally well done, with candidates either using a tree diagram or a binomial approach.

12b. [3 marks]Hence, find .

MarkschemeMETHOD 1

(seen anywhere) (A1)

correct substitution into formula for (A1)

eg ,

A1 N3

METHOD 2

attempt to substitute into (M1)

eg

correct substitution into (A1)

eg

A1 N3

[3 marks]

Examiners reportPart (a)(iii) proved difficult, with many either having trouble finding or using .

12c. [4 marks]

Fourteen balls are drawn from the bag, with replacement.

(b) Find the probability that exactly five of the balls are gold.

(c) Find the probability that at most five of the balls are gold.

MarkschemeLet be the number of gold balls drawn from the bag.

(b) evidence of recognizing binomial (seen anywhere) (M1)

eg ,

A1 N2

[2 marks]

(c) recognize need to find (M1)

A1 N2

[2 marks]

Total [4 marks]

Examiners reportA great majority were confident solving part (b) with the GDC, although some did write the binomial term. Those candidates who didnot use the binomial function on the GDC had more difficulty in part (c), although a pleasing number were still able to identify thatthey were seeking .

12d. [3 marks]Given that at most five of the balls are gold, find the probability that exactly five of the balls are gold. Give the answer correctto two decimal places.

MarkschemeLet be the number of gold balls drawn from the bag.

recognizing conditional probability (M1)

eg , , ,

(A1)

(to dp) A1 N2

[3 marks]

Examiners reportWhile most candidate knew to use conditional probability in part (d), fewer were able to do so successfully, and even fewer stillcorrectly rounded their answer to two decimal places. The most common error was to multiply probabilities to find the intersectionneeded for the conditional probability formula. Overall, candidates seemed better prepared for probability.

13a. [2 marks]

The cumulative frequency curve below represents the marks obtained by 100 students.

Find the median mark.

Examiners reportOverall, this question was done well by candidates. In part (a), a surprising number of candidates found the median position (thecumulative frequency) on the y-axis, but did not find the median mark on the x-axis.

13b. [3 marks]Find the interquartile range.

Markschemelower quartile , upper quartile (A1)(A1)

interquartile range A1 N3

[3 marks]

Examiners reportOverall, this question was done well by candidates. In part (a), a surprising number of candidates found the median position (thecumulative frequency) on the y-axis, but did not find the median mark on the x-axis. Similar misunderstanding was shown by somecandidates in part (b), when attempting to find the interquartile range.

14a. [2 marks]

The random variable X has the following probability distribution, with .

Find the value of r .

Markschemeattempt to substitute (M1)

e.g.

A1 N2

[2 marks]

Examiners reportThe majority of candidates were successful in earning full marks on this question.

14b. [6 marks]Given that , find the value of p and of q .

Markschemecorrect substitution into (seen anywhere) (A1)

e.g.

correct equation A1

e.g. ,

A1 N1

evidence of choosing M1

e.g. ,


, ,

A1 N2

Note: Exception to the FT rule. Award FT marks on an incorrect value of q, even if q is an inappropriate value. Do not award thefinal A mark for an inappropriate value of p.

[6 marks]

Examiners reportIn part (b), a small number of candidates did not use the correct formula for , even though this formula is given in the formulabooklet. There were also a few candidates who incorrectly assumed that , forgetting that the sum of the probabilities must equal1. There were a few candidates who left this question blank, which raises concerns about whether they had been exposed toprobability distributions during the course.

15a. [5 marks]Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

(i) Copy and complete the following tree diagram.

(ii) Find the probability that two white balls are chosen.

Markscheme(i)

A1A1A1 N3

(ii) multiplying along the correct branches (may be seen on diagram) (A1)

e.g.

A1 N2

[5 marks]

Examiners reportPart (a) of this question was answered correctly by the large majority of candidates. There were some who did not follow theinstruction to copy and complete the tree diagram on their separate paper, and simply filled in the blanks on the exam paper.

15b. [5 marks]Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probabilitythat they are both white is .

A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bagB.

Find the probability that the two balls are white.

Markscheme , (seen anywhere) (A1)(A1)


e.g.

correct calculation A1

e.g. ,

A1 N3

[5 marks]

Examiners reportIn part (b), many candidates struggled with finding the compound probability, and did not use the provided information in theappropriate manner. Quite a few candidates seemed to be confused about when they should add the probabilities or when they shouldmultiply.

15c. [4 marks]Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.

Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B, the probabilitythat they are both white is .

A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they are chosen from bagB.

Given that both balls are white, find the probability that they were chosen from bag A.


e.g. ,

correct numerator (A1)

e.g.

correct denominator (A1)

e.g.

probability A1 N3

[4 marks]

Examiners reportIn part (c), many recognized that the question dealt with conditional probability, and many tried to use the formula from theinformation booklet, but failed to realize that they had already found the required values for the numerator and denominator in theirworking for part (b).

Throughout this question, it was discouraging to see the large number of candidates making arithmetic errors. There were a surprisingnumber of candidates who multiplied fractions incorrectly, or found an incorrect value for simple multiplication such as or

.

16a. [3 marks]The heights of a group of seven-year-old children are normally distributed with mean and standard deviation . Achild is chosen at random from the group.

Find the probability that this child is taller than .

Markschemeevidence of appropriate method (M1)

e.g. , sketch of normal curve showing mean and ,

(A1)

A1 N3

[3 marks]

Examiners reportThere were many completely successful attempts at this question, with good use of formulae and calculator features.

16b. [3 marks]The heights of a group of seven-year-old children are normally distributed with mean and standard deviation . Achild is chosen at random from the group.

The probability that this child is shorter than is . Find the value of k .

Markscheme (A1)

set up equation (M1)

e.g. , sketch

A1 N3

[3 marks]

Examiners reportThere were many completely successful attempts at this question, with good use of formulae and calculator features.

However, in part (b) some candidates did not recognize the need to find the standardized value and set their equation equal to theprobability given in the question, thus earning only one mark.

17a. [4 marks]A factory makes lamps. The probability that a lamp is defective is 0.05. A random sample of 30 lamps is tested.

Find the probability that there is at least one defective lamp in the sample.

Markschemeevidence of recognizing binomial (seen anywhere) (M1)

e.g. ,

finding (A1)


e.g. complement, summing probabilities

probability is A1 N3

[4 marks]

Examiners reportAlthough candidates seemed more confident in attempting binomial probabilities than in previous years, some of them failed torecognize the binomial nature of the question in part (a). Many knew that the complement was required, but often used or instead of .

17b. [4 marks]A factory makes lamps. The probability that a lamp is defective is 0.05. A random sample of 30 lamps is tested.

Given that there is at least one defective lamp in the sample, find the probability that there are at most two defective lamps.

Markschemeidentifying correct outcomes (seen anywhere) (A1)

e.g. , 1 or 2 defective,

recognizing conditional probability (seen anywhere) R1

e.g. , , P(at most 2|at least 1)

appropriate approach involving conditional probability (M1)

e.g. , ,

probability is A1 N2

[4 marks]

Examiners reportPart (b) was poorly answered. While some candidates recognized that it was a conditional probability, very few were able to correctlyapply the formula, identify the outcomes and follow on to achieve the correct result.

Only a few could find the intersection of the events correctly. Several thought the numerator was a product (i.e. ), and then cancelled common factors with the denominator. Others realized that and were

required but multiplied their probabilities.

This was the most commonly missed out question from Section A.

18a. [2 marks]

The ages of people attending a music concert are given in the table below.

Find p .

Markschemeevidence of valid approach (M1)

e.g. , line on graph at

A1 N2

[2 marks]

Examiners reportPart (a) was generally answered correctly, with most candidates showing a good grasp of cumulative frequency from a table.

18b. [5 marks]The cumulative frequency diagram is given below.

Use the diagram to estimate

(i) the 80th percentile;

(ii) the interquartile range.

Markscheme(i) evidence of valid approach (M1)

e.g. line on graph, , using complement

A1 N2

(ii) ; (A1)(A1)

(accept any notation that suggests an interval) A1 N3

[5 marks]

Examiners reportA surprising number of candidates had difficulty reading values off the cumulative frequency curve. A common incorrect answer for(b)(i) was 29, indicating carelessness with the given scale. Too many candidates gave 40 and 120 for the quartile values.

19a. [1 mark]

Events A and B are such that , and .

The values q , r , s and t represent probabilities.

Write down the value of t .

Markscheme A1 N1

[1 mark]

Examiners reportParts (a), (b), and (c)(i) of this Venn diagram probability question were answered quite well with candidates consistently earning fullmarks.

19b. [3 marks](i) Show that .

(ii) Write down the value of q and of s .

Markscheme(i) correct values A1

e.g. ,

AG N0

(ii) , A1A1 N2

[3 marks]

Examiners reportParts (a), (b), and (c)(i) of this Venn diagram probability question were answered quite well with candidates consistently earning fullmarks. Only a few candidates worked backwards from the given in the "show that" portion of part (b).

19c. [3 marks](i) Write down .

(ii) Find .

Markscheme(i) A1 N1

(ii) A2 N2

[3 marks]

Examiners reportMany candidates struggled on part (c)(ii), either not recognizing conditional probability or multiplying probabilities to find thenumerator as if the events were independent. A number of candidates who successfully found the probability in part (c)(ii) left theirincomplete answer of .

20a. [4 marks]The probability of obtaining “tails” when a biased coin is tossed is . The coin is tossed ten times. Find the probability ofobtaining at least four tails.

Markschemeevidence of recognizing binomial distribution (M1)

e.g. , ,

EITHER

(A1)

evidence of using complement (M1)

e.g. any probability,

A1 N3

OR

summing the probabilities from to (M1)

correct expression or values (A1)

e.g. ,

0.919424

A1 N3

[4 marks]

Examiners reportThis was an accessible problem that created some difficulties for candidates. Most were able to recognize the binomial nature of theproblem but were confused by the phrase "at least four tails" which was often interpreted as the complement of four or less. Pooralgebraic manipulation also led to unnecessary errors that the calculator approach would have avoided.

20b. [3 marks]The probability of obtaining “tails” when a biased coin is tossed is 0.57. The coin is tossed ten times. Find the probability ofobtaining the fourth tail on the tenth toss.


e.g. three tails in nine tosses,

correct calculation

e.g. , (A1)

A1 N2

[3 marks]

Examiners reportThis was an accessible problem that created some difficulties for candidates. Most were able to recognize the binomial nature of theproblem but were confused by the phrase "at least four tails" which was often interpreted as the complement of four or less. Pooralgebraic manipulation also led to unnecessary errors that the calculator approach would have avoided.

21a. [3 marks]

The histogram below shows the time T seconds taken by 93 children to solve a puzzle.

The following is the frequency distribution for T .

(i) Write down the value of p and of q .

(ii) Write down the median class.

Markscheme(i) , A1A1 N2

(ii) A1 N1

[3 marks]

Examiners reportParts (a) and (b) were generally well done. The terms "median" and "median class" were often confused.

21b. [2 marks]A child is selected at random. Find the probability that the child takes less than 95 seconds to solve the puzzle.


e.g. adding frequencies

A1 N2

[2 marks]

Examiners reportParts (a) and (b) were generally well done. The terms "median" and "median class" were often confused.

21c. [2 marks]Consider the class interval .

(i) Write down the interval width.

(ii) Write down the mid-interval value.

Markscheme(i) 10 A1 N1

(ii) 50 A1 N1

[2 marks]

Examiners reportIn part (c) some candidates had problems with the term "interval width" and there were some rather interesting mid-interval valuesnoted.

21d. [4 marks]Hence find an estimate for the

(i) mean;

(ii) standard deviation.

Markscheme(i) evidence of approach using mid-interval values (may be seen in part (ii)) (M1)

A2 N3

(ii)

A1 N1

[4 marks]

Examiners reportIn part (d), candidates often ignored the "hence" command and estimated values from the graph rather than from the information inpart (c).

21e. [2 marks]John assumes that T is normally distributed and uses this to estimate the probability that a child takes less than 95 seconds tosolve the puzzle.

Find John’s estimate.

Markschemee.g. standardizing,

A1 N2

[2 marks]

Examiners reportThose who correctly obtained the mean and standard deviation had little difficulty with part (e) although candidates often usedunfamiliar calculator notation as their working or used the mid-interval value as the mean of the distribution.

22a. [1 mark]

A box contains six red marbles and two blue marbles. Anna selects a marble from the box. She replaces the marble and then selects a secondmarble.

Write down the probability that the first marble Anna selects is red.

MarkschemeNote: In this question, method marks may be awarded for selecting without replacement, as noted in the examples.

A1 N1

[1 mark]

Examiners reportCandidates did very well on parts (a) and (b) of this probability question.

22b. [2 marks]Find the probability that Anna selects two red marbles.

Markschemeattempt to find (M1)

e.g. , ,

A1 N2

[2 marks]

Examiners reportCandidates did very well on parts (a) and (b) of this probability question, and knew to multiply the probabilities of independent eventsin part (b). However, in part (c), very few candidates considered that there are two ways to draw one red and one blue marble, andtherefore did not earn full marks on this question. There were also some candidates who tried to add, rather than multiply, theprobabilities in parts (b) and (c).

22c. [3 marks]Find the probability that one marble is red and one marble is blue.

MarkschemeMETHOD 1

attempt to find (M1)

e.g. , ,

recognizing two ways to get one red, one blue (M1)

e.g. , ,

A1 N2

[3 marks]

METHOD 2

recognizing that is (M1)

attempt to find and (M1)

e.g. , ; ,

A1 N2

[3 marks]

Examiners reportCandidates did very well on parts (a) and (b) of this probability question, and knew to multiply the probabilities of independent eventsin part (b). However, in part (c), very few candidates considered that there are two ways to draw one red and one blue marble, andtherefore did not earn full marks on this question. There were also some candidates who tried to add, rather than multiply, theprobabilities in parts (b) and (c).

23. [6 marks]The random variable X has the following probability distribution.

Given that , find q .

Markschemecorrect substitution into (seen anywhere) A1

e.g. ,

recognizing (seen anywhere) (M1)

correct substitution into A1

e.g.

attempt to solve simultaneous equations (M1)


e.g. ,

A1 N4

[6 marks]

Examiners reportCandidates generally earned either full marks or only one mark on this question. The most common error was where candidates onlywrote the equation for , and tried to rearrange that equation to solve for q. The candidates who also knew that the sum ofthe probabilities must be equal to 1 were very successful in solving the resulting system of equations.

24a. [4 marks]

Let , where .

Find the values of k such that has two equal roots.

MarkschemeMETHOD 1

evidence of discriminant (M1)

e.g. , discriminant = 0

correct substitution into discriminant A1

e.g. ,

A1A1 N3

METHOD 2

recognizing that equal roots means perfect square (R1)

e.g. attempt to complete the square,

correct working

e.g. , A1

A1A1 N3

[4 marks]

Examiners reportA good number of candidates were successful in using the discriminant to find the correct values of k in part (a), however, there weremany who tried to use the quadratic formula without recognizing the significance of the discriminant.

24b. [4 marks]Each value of k is equally likely for . Find the probability that has no roots.


e.g.

correct working for k A1

e.g. , , list all correct values of k

A2 N3

[4 marks]

Examiners reportPart (b) was very poorly done by nearly all candidates. Common errors included finding the wrong values for k, and not realizing thatthere were 11 possible values for k.

25a. [1 mark]

The cumulative frequency curve below represents the heights of 200 sixteen-year-old boys.

Use the graph to answer the following.

Write down the median value.

Markscheme A1 N1

[1 mark]

Examiners reportParts (a) and (b) were generally well done.

25b. [2 marks]A boy is chosen at random. Find the probability that he is shorter than .

Markschemeattempt to find number shorter than 161 (M1)

e.g. line on graph, 12 boys

A1 N2

[2 marks]

Examiners reportParts (a) and (b) were generally well done.

25c. [3 marks]Given that of the boys are taller than , find h .

MarkschemeMETHOD 1

have a height less than h (A1)

(36 may be seen as a line on the graph) (A1)

(cm) A1 N2

METHOD 2

(164 may be seen as a line on the graph) (A1)

(A1)

(cm) A1 N2

[3 marks]

Examiners reportSome candidates could only earn the first mark in part (c) for finding of 200. Others gave the answer as 164, neglecting tosubtract this value from the total of 200.

26a. [2 marks]

A company produces a large number of water containers. Each container has two parts, a bottle and a cap. The bottles and caps are tested tocheck that they are not defective.

A cap has a probability of 0.012 of being defective. A random sample of 10 caps is selected for inspection.

Find the probability that exactly one cap in the sample will be defective.

MarkschemeNote: There may be slight differences in answers, depending on whether candidates use tables or GDCs, or their 3 sf answers insubsequent parts. Do not penalise answers that are consistent with their working and check carefully for FT.

evidence of recognizing binomial (seen anywhere in the question) (M1)

e.g. , ,

A1 N2

[2 marks]

Examiners reportMany stronger candidates were completely successful with this question, employing technology efficiently. A number of candidatesdid not recognize the binomial probability in parts (a) and (b), and in part (b) a proportion of candidates just subtracted their part (a)answer from one. Candidates had more success with the normal distribution and many obtained follow-through marks in part (e) afteran error made in part (b). Many candidates did not appreciate the independence in part (e) and added probabilities rather thanmultiplying them. A number of candidates were penalised for not giving their answers to 3 significant figures.

26b. [2 marks]The sample of caps passes inspection if at most one cap is defective. Find the probability that the sample passes inspection.


e.g. ,

A1 N2

[2 marks]


26c. [5 marks]The heights of the bottles are normally distributed with a mean of and a standard deviation of .

(i) Copy and complete the following diagram, shading the region representing where the heights are less than .

(ii) Find the probability that the height of a bottle is less than .

Markscheme(i)

A1A1 N2

Note: Award A1 for vertical line to right of mean, A1 for shading to left of their vertical line.

(ii) valid approach (M1)

e.g.

working to find standardized value (A1)

e.g. , 2.1

A1 N3

[5 marks]


26d. [5 marks](i) A bottle is accepted if its height lies between and . Find the probability that a bottle selected at randomis accepted.

(ii) A sample of 10 bottles passes inspection if all of the bottles in the sample are accepted. Find the probability that the sample passesinspection.


e.g. ,


e.g.

A1 N3

(ii) correct working (A1)

e.g. ,

(accept 0.694 from tables) A1 N2

[5 marks]


26e. [2 marks]The bottles and caps are manufactured separately. A sample of 10 bottles and a sample of 10 caps are randomly selected fortesting. Find the probability that both samples pass inspection.


e.g. ,

(accept from tables) A1 N2

[2 marks]


27a. [1 mark]

The probability distribution of a discrete random variable X is given by

.

Write down .

Markscheme A1 N1

[1 mark]

Examiners reportAlthough many candidates were successful in working with the probability function, students had difficulty following the "show that"instruction of this question. Many substituted and worked backwards to show that the sum of probabilities is 1. Some wouldargue that does not work, but were unable to give a complete justification for . A good number of students seemedunprepared to find an expected value. Many candidates wrote a formula and did not know what to do with it, while others divided

by 3 or by 6, which confuses the concept of a mean in a probability distribution with the more common understanding.

27b. [4 marks]Show that .

Markscheme (A1)

(A1)

setting the sum of probabilities M1

e.g. ,

(accept ) A1

AG N0

[4 marks]

Examiners reportAlthough many candidates were successful in working with the probability function, students had difficulty following the "show that"instruction of this question. Many substituted and worked backwards to show that the sum of probabilities is 1. Some wouldargue that does not work, but were unable to give a complete justification for .

27c. [2 marks]Find .

Markschemecorrect substitution into A1

e.g.

A1 N1

[2 marks]

Examiners reportA good number of students seemed unprepared to find an expected value. Many candidates wrote a formula and did not know whatto do with it, while others divided by 3 or by 6, which confuses the concept of a mean in a probability distribution with themore common understanding.

28a. [5 marks]

In a group of 16 students, 12 take art and 8 take music. One student takes neither art nor music. The Venn diagram below shows the events artand music. The values p , q , r and s represent numbers of students.

(i) Write down the value of s .

(ii) Find the value of q .

(iii) Write down the value of p and of r .

Markscheme(i) A1 N1

(ii) evidence of appropriate approach (M1)

e.g. ,

A1 N2

(iii) , A1A1 N2

[5 marks]

Examiners reportA majority of candidates found the values in the Venn diagram easily. Common errors include giving , and also neglecting s infinding (e.g. ) . Some interpreted the values as probabilities, despite the question explicitly stating that p, q, r and srepresent numbers of students. Occasionally the values for p and r were misinterpreted as being inclusive of q. Follow-through markswere often earned in subsequent parts for such cases.

28b. [4 marks](i) A student is selected at random. Given that the student takes music, write down the probability the student takes art.

(ii) Hence, show that taking music and taking art are not independent events.

Markscheme(i) A2 N2

(ii) METHOD 1

A1

evidence of correct reasoning R1

e.g.

the events are not independent AG N0

METHOD 2

A1

evidence of correct reasoning R1

e.g.

the events are not independent AG N0

[4 marks]

Examiners reportFor (b), rather than think of the situations conceptually, most candidates reached for the formula for conditional probability, withmixed results. Few candidates considered that independence means . Most applied , withmany giving incomplete or incorrect calculations. Some candidates compared the wrong things and showed, for example, that ,which incorrectly compares with . Others stated that because there is an intersection, the events are independent,which is an insufficient explanation.

28c. [4 marks]Two students are selected at random, one after the other. Find the probability that the first student takes only music and thesecond student takes only art.

Markscheme (seen anywhere) A1

(seen anywhere) A1

evidence of valid approach (M1)

e.g.

A1 N2

[4 marks]

Examiners reportPart (c) was commonly answered as if there is replacement, with many candidates calculating . However, implicit in thephrasing "one after the other" is that there is no replacement.

29a. [3 marks]

A random variable X is distributed normally with a mean of 20 and variance 9.

Find .

Markscheme (A1)

evidence of attempt to find (M1)

e.g. ,

A1 N3

[3 marks]

Examiners reportThis question clearly demonstrated that some centres are still not giving adequate treatment to this topic. A great many candidatesneglected to find the standard deviation and used the variance throughout. More still did not leave their answers to the requiredaccuracy. Ignoring the use of the variance, responses to part (a) demonstrated that most candidates were comfortable finding therequired probability using their calculator or setting up a suitable standardized equation.

29b. [5 marks]Let .

(i) Represent this information on the following diagram.

(ii) Find the value of k .

Markscheme

A1A1 N2

Note: Award A1 with shading that clearly extends to right of the mean, A1 for any correct label, either k, area or their value of k.

(ii) (A1)

attempt to set up an equation (M1)

e.g. ,

A1 N3

[5 marks]

Examiners reportThis question clearly demonstrated that some centres are still not giving adequate treatment to this topic. A great many candidatesneglected to find the standard deviation and used the variance throughout. More still did not leave their answers to the requiredaccuracy. Ignoring the use of the variance, responses to part (a) demonstrated that most candidates were comfortable finding therequired probability using their calculator or setting up a suitable standardized equation. In part (b) (i), the sketch was often poorlyshaded or incorrectly labelled. In (b) (ii), candidates frequently confused the z-score with the given probability of 0.85. Calculatorapproaches were more successful than working by hand but candidates should remember to avoid the use of calculator notation intheir working, as it is not correct mathematical notation.

30a. [2 marks]

A box holds 240 eggs. The probability that an egg is brown is 0.05.

Find the expected number of brown eggs in the box.

Markschemecorrect substitution into formula for (A1)

e.g.

A1 N2

[2 marks]

Examiners reportPart (a) was answered correctly by most candidates.

30b. [2 marks]Find the probability that there are 15 brown eggs in the box.

Markschemeevidence of recognizing binomial probability (may be seen in part (a)) (M1)

e.g. ,

A1 N2

[2 marks]

Examiners reportIn parts (b) and (c), many failed to recognize the binomial nature of this experiment and opted for incorrect techniques in simpleprobability.

30c. [3 marks]Find the probability that there are at least 10 brown eggs in the box.

Markscheme (A1)


e.g. using complement, summing probabilities

A1 N3

[3 marks]

Examiners reportIn parts (b) and (c), many failed to recognize the binomial nature of this experiment and opted for incorrect techniques in simpleprobability. Although several candidates appreciated that (c) involved the idea of a complement, some resorted to elaborateprobability addition suggesting they were unaware of the capabilities of their GDC. There was also a great deal of evidence tosuggest that candidates did not understand the phrase "at least 10" as several candidates found either ,

or .

31a. [3 marks]

A company uses two machines, A and B, to make boxes. Machine A makes of the boxes.

of the boxes made by machine A pass inspection.

of the boxes made by machine B pass inspection.

A box is selected at random.

Find the probability that it passes inspection.

Markschemeevidence of valid approach involving A and B (M1)

e.g. , tree diagram

correct expression (A1)

e.g.

A1 N2

[3 marks]

Examiners reportPart (a) was usually well done. Those candidates that did not succeed with this part often did not show a correct tree diagramindicating that they did not really understand the problem or indeed how to start it.

31b. [4 marks]The company would like the probability that a box passes inspection to be 0.87.

Find the percentage of boxes that should be made by machine B to achieve this.

Markschemeevidence of recognizing complement (seen anywhere) (M1)

e.g. , , , ,


e.g. ,

correct expression A1

e.g. , ,

from B A1 N2

[4 marks]

Examiners reportMany successful attempts to (b) relied on "guess and check" or intuitive solutions while a surprising number of candidates could notmanage to systematically set up an appropriate algebraic expression involving a complement.

32a. [4 marks]

The Venn diagram below shows events A and B where , and . The values m , n , p and q areprobabilities.

(i) Write down the value of n .

(ii) Find the value of m , of p , and of q .

Markscheme(i) A1 N1

(ii) , , A1A1A1 N3

[4 marks]

Examiners reportMost candidates were able to find the correct values for the Venn diagram. Unfortunately, however, there were many candidates whodid not understand what each region of the diagram represents. For example, a very common error was thinking that ,rather than the correct .

32b. [2 marks]Find .

Markschemeappropriate approach

e.g. , , (M1)

A1 N2

[2 marks]

Examiners reportCandidates seemed to understand the idea of the complement in part (b), but some were not able to find the correct answer because ofconfusion over the separation of the different regions in the diagram.

33a. [3 marks]

A scientist has 100 female fish and 100 male fish. She measures their lengths to the nearest cm. These are shown in the following box andwhisker diagrams.

Find the range of the lengths of all 200 fish.

Markschemecorrect end points (A1)(A1)

max = 27 , min = 4

range = 23 A1 N3

[3 marks]

Examiners reportWhile there were a large number of candidates who answered both parts of this question correctly, a surprising number did not knowhow to find the range of all 200 fish in part (a). Common errors included finding the ranges of the male and female fish separately, oraveraging the separate ranges of the male and female fish.

33b. [2 marks]Four cumulative frequency graphs are shown below.

Which graph is the best representation of the lengths of the female fish?

MarkschemeGraph 3 A2 N2

[2 marks]

Examiners reportSome candidates did not interpret the cumulative frequency graphs correctly, or just seemed to guess which graph was correct. Themost common incorrect "guess" was graph 4, likely because this graph had a more familiar cumulative shape.

34a. [2 marks]

Let the random variable X be normally distributed with mean 25, as shown in the following diagram.

The shaded region between 25 and 27 represents of the distribution.

Find .

Markschemesymmetry of normal curve (M1)

e.g.

A1 N2

[2 marks]

Examiners reportThis question proved challenging for many candidates. A surprising number did not use the symmetry of the normal curve to find theprobability required in (a). While many students were able to set up a standardized equation in (b), far fewer were able to use thecomplement to find the correct z-score. Others used 0.8 as the z-score. A common confusion when approaching parts (a) and (b) waswhether to use a probability or a z-score. Additionally, many candidates seemed unsure of appropriate notation on this problem whichwould have allowed them to better demonstrate their method.

34b. [5 marks]Find the standard deviation of X .

MarkschemeMETHOD 1

finding standardized value (A1)

e.g.

evidence of complement (M1)

e.g. , , 0.8

finding z-score (A1)

e.g.

attempt to set up equation involving the standardized value M1

e.g. ,

A1 N3

METHOD 2

set up using normal CDF function and probability (M1)

e.g. ,

correct equation A2

e.g. ,

attempt to solve the equation using GDC (M1)

e.g. solver, graph, trial and error (more than two trials must be shown)

A1 N3

[5 marks]

Examiners reportThis question proved challenging for many candidates. A surprising number did not use the symmetry of the normal curve to find theprobability required in (a). While many students were able to set up a standardized equation in (b), far fewer were able to use thecomplement to find the correct z-score. Others used 0.8 as the z-score. A common confusion when approaching parts (a) and (b) waswhether to use a probability or a z-score. Additionally, many candidates seemed unsure of appropriate notation on this problem whichwould have allowed them to better demonstrate their method.

35a. [3 marks]

Two fair 4-sided dice, one red and one green, are thrown. For each die, the faces are labelled 1, 2, 3, 4. The score for each die is the numberwhich lands face down.

List the pairs of scores that give a sum of 6.

Markschemethree correct pairs A1A1A1 N3

e.g. (2, 4), (3, 3), (4, 2) , R2G4, R3G3, R4G2

[3 marks]

Examiners reportAll but the weakest candidates managed to score full marks for parts (a) and (b). An occasional error in part (a) was includingadditional pair(s) or listing (3, 3) twice.

35b. [3 marks]The probability distribution for the sum of the scores on the two dice is shown below.

Find the value of p , of q , and of r .

Markscheme , , A1A1A1 N3

[3 marks]

Examiners reportAll but the weakest candidates managed to score full marks for parts (a) and (b).

35c. [6 marks]Fred plays a game. He throws two fair 4-sided dice four times. He wins a prize if the sum is 5 on three or more throws.

Find the probability that Fred wins a prize.

Markschemelet X be the number of times the sum of the dice is 5


e.g. , tree diagram, 5 sets of outcomes produce a win

one correct parameter (A1)

e.g. , ,

Fred wins prize is (A1)

appropriate approach to find probability M1

e.g. complement, summing probabilities, using a CDF function


e.g. , , ,

A1 N3

[6 marks]

Examiners reportMany candidates found part (c) challenging, as they failed to recognize the binomial probability. Successful candidates generally usedeither the binomial CDF function or the sum of two binomial probabilities. Some used approaches like multiplying probabilities ortree diagrams, but these were less successful.

36a. [1 mark]

The diagram below shows the probabilities for events A and B , with .

Write down the value of p .

Markscheme A1 N1

[1 mark]

Examiners reportWhile nearly every candidate answered part (a) correctly, many had trouble with the other parts of this question.


Markschememultiplying along the branches (M1)

e.g. ,

adding products of probabilities of two mutually exclusive paths (M1)

e.g. ,

A1 N2

[3 marks]

Examiners reportIn part (b), many candidates did not multiply along the branches of the tree diagram to find the required values, and many did notrealize that there were two paths for P(B). There were also many candidates who understood what the question required, but then didnot know how to multiply fractions correctly, and these calculation errors led to an incorrect answer.


Markschemeappropriate approach which must include (may be seen on diagram) (M1)

e.g. (do not accept )

(A1)

A1 N2

[3 marks]

Examiners reportIn part (c), most candidates attempted to use a formula for conditional probability found in the information booklet, but very fewsubstituted the correct values.

37a. [3 marks]

Consider the events A and B, where , and .

The Venn diagram below shows the events A and B, and the probabilities p, q and r.

Write down the value of

(i) p ;

(ii) q ;

(iii) r.

Markscheme(i) A1 N1

(ii) A1 N1

(iii) A1 N1

[3 marks]

Examiners reportAs the definitions of p and q were not clear to candidates, both responses of p , q and p , q were accepted forfull marks. However, finding r eluded many.

37b. [2 marks]Find the value of .

Markscheme A2 N2

Note: Award A1 for an unfinished answer such as .

[2 marks]

Examiners reportFew candidates answered the conditional probability correctly. Many attempted to use the formula in the booklet without consideringthe complement, and there was little evidence of the Venn diagram being utilized as a helpful aid.

37c. [1 mark]Hence, or otherwise, show that the events A and B are not independent.

Markschemevalid reason R1

e.g. ,

thus, A and B are not independent AG N0

[1 mark]

Examiners reportTo show the events are not independent, many correctly reasoned that . A handful recognized that is analternative approach that uses the answer in part (b). Some candidates do not know the difference between independent andmutually exclusive.

38a. [4 marks]

José travels to school on a bus. On any day, the probability that José will miss the bus is .

If he misses his bus, the probability that he will be late for school is .

If he does not miss his bus, the probability that he will be late is .

Let E be the event “he misses his bus” and F the event “he is late for school”.

The information above is shown on the following tree diagram.

Find

(i) ;

(ii) .

Markscheme(i) A1 N1

(ii) evidence of multiplying along the branches (M1)

e.g. ,

adding probabilities of two mutually exclusive paths (M1)

e.g. ,

A1 N2

[4 marks]

Examiners reportCandidates generally handled some or all of parts (a) and (b) well. Errors included adding probabilities along branches and trying touse the union formula from the information booklet.

38b. [5 marks]Find the probability that

(i) José misses his bus and is not late for school;

(ii) José missed his bus, given that he is late for school.

Markscheme(i) (A1)

A1

(ii) recognizing this is (M1)

e.g.

A2 N3

[5 marks]

Examiners reportCandidates generally handled some or all of parts (a) and (b) well. Errors included adding probabilities along branches and trying touse the union formula from the information booklet. On part (b)(ii), many candidates knew that they were supposed to use some typeof conditional probability but did not know how to find . Many candidates made errors working with fractions. Somecandidates who missed part (a)(ii) were able to earn follow-through credit on part (b)(ii).

38c. [3 marks]The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and Tuesday.

Copy and complete the probability distribution table.

Markscheme

A2A1 N3

[3 marks]

Examiners reportMany candidates had difficulty completing the probability distribution table. While the common error of finding the probability for

as was understandable as the candidate did not appreciate that there were two ways of paying three euros, it wasdisappointing that these candidates often correctly found as and did not note that the probabilities failed to sum to one.These candidates could not earn full follow-through marks on their expected value calculation in part (d). Some candidates did use theprobabilities summing to one with incorrect probabilities in part (c); these candidates often earned full follow-through marks in part(d), as a majority of candidates knew the method for finding expected value.

38d. [2 marks]The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and Tuesday.

Find the expected cost for José for both days.

Markschemecorrect substitution into formula (M1)

e.g. ,

(euros) A1 N2

[2 marks]

Examiners reportMany candidates had difficulty completing the probability distribution table. While the common error of finding the probability for

as was understandable as the candidate did not appreciate that there were two ways of paying three Euros, it wasdisappointing that these candidates often correctly found as and did not note that the probabilities failed to sum to one.These candidates could not earn full follow-through marks on their expected value calculation in part (d). Some candidates did use theprobabilities summing to one with incorrect probabilities in part (c); these candidates often earned full follow-through marks in part(d), as a majority of candidates knew the method for finding expected value.

39a. [2 marks]

A standard die is rolled 36 times. The results are shown in the following table.

Write down the standard deviation.

Markscheme A2 N2

[2 marks]

Examiners reportSurprisingly, this question was not answered well primarily due to incorrect GDC use and a lack of understanding of the terms"median" and "interquartile range". Many candidates opted for an analytical approach in part (a) which always resulted in mistakes.

39b. [1 mark]Write down the median score.

Markschememedian A1 N1

[1 mark]

Examiners reportSome candidates wrote the down the mean instead of the median in part (b).

39c. [3 marks]Find the interquartile range.

Markscheme , (may be seen in a box plot) (A1)(A1)

(accept any notation that suggests the interval 3 to 5) A1 N3

[3 marks]

Examiners reportSurprisingly, this question was not answered well primarily due to incorrect GDC use and a lack of understanding of the terms"median" and "interquartile range".

40a. [6 marks]

A test has five questions. To pass the test, at least three of the questions must be answered correctly.

The probability that Mark answers a question correctly is . Let X be the number of questions that Mark answers correctly.

(i) Find E(X ) .

(ii) Find the probability that Mark passes the test.

Markscheme(i) valid approach (M1)

e.g. ,

A1 N2

(ii) evidence of appropriate approach involving binomial (M1)

e.g.

recognizing that Mark needs to answer 3 or more questions correctly (A1)

e.g.

valid approach M1

e.g. ,

A1 N3

[6 marks]

Examiners reportThere was wide spectrum of success on this problem. Candidates could normally find E(X) using but many failed to recognizethat the "experiment" was binomial or that for Mark to the pass the test, he needed to answer either 3, 4 or 5 questions correctly.

40b. [8 marks]

Bill also takes the test. Let Y be the number of questions that Bill answers correctly.

The following table is the probability distribution for Y .

(i) Show that .

(ii) Given that , find a and b .

Markscheme(i) evidence of summing probabilities to 1 (M1)

e.g.

some simplification that clearly leads to required answer

e.g. A1

AG N0

(ii) correct substitution into the formula for expected value (A1)

e.g.

some simplification (A1)

e.g.

correct equation A1

e.g.

evidence of solving (M1)

, A1A1 N4

[8 marks]

Examiners reportPart (b) was generally well done although there were a number of algebraic errors particularly in part (b) (ii), leading to incorrectvalues of a and b. Again, appropriate use of the GDC here would have eliminated these errors.

40c. [3 marks]Find which student is more likely to pass the test.

Markschemeattempt to find probability Bill passes (M1)

e.g.

correct value 0.19 A1

Bill (is more likely to pass) A1 N0

[3 marks]

Examiners reportIn (c), candidates had trouble with the command term, "find" and often just wrote down either "Mark" or "Bill".

41a. [4 marks]

The following frequency distribution of marks has mean 4.5.

Find the value of x.

Markscheme , (seen anywhere) A1

evidence of substituting into mean (M1)

correct equation A1

e.g. ,

A1 N2

[4 marks]

Examiners reportSurprisingly, this question was not well done by many candidates. A good number of candidates understood the importance of thefrequencies in calculating mean. Some neglected to sum the frequencies for the denominator, which often led to a negative value for afrequency. Unfortunately, candidates did not appreciate the unreasonableness of this result.

41b. [2 marks]Write down the standard deviation.

Markscheme A2 N2

[2 marks]

Examiners reportSurprisingly, this question was not well done by many candidates. A good number of candidates understood the importance of thefrequencies in calculating mean. Some neglected to sum the frequencies for the denominator, which often led to a negative value for afrequency. Unfortunately, candidates did not appreciate the unreasonableness of this result. In part (b), many candidates could notfind the standard deviation in their GDC, often trying to calculate it by hand with no success. Further, many could not distinguishbetween the sample and the population standard deviation given in the GDC.

42a. [2 marks]

Evan likes to play two games of chance, A and B.

For game A, the probability that Evan wins is 0.9. He plays game A seven times.

Find the probability that he wins exactly four games.

Markschemeevidence of recognizing binomial probability (may be seen in (b) or (c)) (M1)

e.g. probability , , complementary probabilities

probability A1 N2

[2 marks]

Examiners reportParts of this question were handled very well by a great many candidates. Most were able to recognize the binomial condition andhad little difficulty with part (a). However, more than a few reported the answer as 0.23, thus incurring the accuracy penalty.

42b. [2 marks]

For game B, the probability that Evan wins is p . He plays game B seven times.

Write down an expression, in terms of p , for the probability that he wins exactly four games.

Markschemecorrect expression A1A1 N2

e.g. ,

Note: Award A1 for binomial coefficient (accept ) , A1 for .

[2 marks]

Examiners reportThose candidates that were successful in part (a) could easily write the required expression for part (b).

42c. [3 marks]Hence, find the values of p such that the probability that he wins exactly four games is 0.15.

Markschemeevidence of attempting to solve their equation (M1)

e.g. , sketch

, A1A1 N3

[3 marks]

Examiners reportIn part (c), many candidates set up the question correctly or set their expression from (b) equal to 0.15, however few candidatesconsidered the GDC as a method to solve the equation. Rather, those who attempted usually tried to expand the polynomial, and stilldid not use the GDC to solve this equation. A graphical approach to the solution would reveal that there are two solutions for p, butfew caught this subtlety.

43a. [2 marks]

The weights of players in a sports league are normally distributed with a mean of , (correct to three significant figures). It is known that of the players have weights between and . The probability that a player weighs less than is 0.05.

Find the probability that a player weighs more than .


e.g. , diagram showing values in a normal curve

A1 N2

[2 marks]

Examiners reportThis question was quite accessible to those candidates in centres where this topic is given the attention that it deserves. Mostcandidates handled part (a) well using the basic properties of a normal distribution.

43b. [4 marks](i) Write down the standardized value, z, for .

(ii) Hence, find the standard deviation of weights.

Markscheme(i) A1 N1


e.g. ,

correct substitution A1

e.g.

A1 N1

[4 marks]

Examiners reportIn part (b) (i), candidates often confused the z-score with the area in the table which led to a standard deviation that was less than zeroin part (b) (ii). At this point, candidates “fudged” results in order to continue with the remaining parts of the question. In (b) (ii), the“hence” command was used expecting candidates to use the results of (b) (i) to find a standard deviation of 4.86. Unfortunately, manydecided to use their answers and the information from part (a) resulting in quite a different standard deviation of 5.79. Recognizingthe inconsistency in the question, full marks were awarded for this approach, as well as full follow-through in subsequent parts of thequestion.

43c. [5 marks]To take part in a tournament, a player’s weight must be within 1.5 standard deviations of the mean.

(i) Find the set of all possible weights of players that take part in the tournament.

(ii) A player is selected at random. Find the probability that the player takes part in the tournament.

Markscheme(i) A1A1A1 N3

Note: Award A1 for 68.8, A1 for 84.4, A1 for giving answer as an interval.


e.g. ,

A1 N2

[5 marks]

Examiners reportCandidates could obtain full marks easily in part (c) with little understanding of a normal distribution but they often confused z-scoreswith data values, adding and subtracting 1.5 from the mean of 76.

43d. [4 marks]Of the players in the league, are women. Of the women, take part in the tournament.

Given that a player selected at random takes part in the tournament, find the probability that the selected player is a woman.


e.g.

(A1)

A1

A1

[4 marks]

Examiners reportIn part (d), few recognized the conditional nature of the question and only determined the probability that a woman qualifies andtakes part in the tournament.

44a. [4 marks]

The following table gives the examination grades for 120 students.

Find the value of

(i) p ;

(ii) q .

Markscheme(a) (i) evidence of appropriate approach (M1)

e.g. ,

A1 N2

(ii) evidence of valid approach (M1)

e.g. their value of p,

A1 N2

[4 marks]

Examiners reportThe majority of candidates had little trouble finding the missing values in the frequency distribution table.

44b. [2 marks]Find the mean grade.


e.g. substituting into , division by 120

mean A1 N2

[2 marks]

Examiners reportMany did not seem comfortable calculating the mean and standard deviation using their GDCs.

The correct mean was often found without the use of the statistical functions on the graphing calculator, but a large number ofcandidates were unable to find the standard deviation.

44c. [1 mark]Write down the standard deviation.

Markscheme1.09 A1 N1

[1 mark]

Examiners reportMany did not seem comfortable calculating the mean and standard deviation using their GDCs.

The correct mean was often found without the use of the statistical functions on the graphing calculator, but a large number ofcandidates were unable to find the standard deviation.

45a. [3 marks]

Jan plays a game where she tosses two fair six-sided dice. She wins a prize if the sum of her scores is 5.

Jan tosses the two dice once. Find the probability that she wins a prize.

Markscheme36 outcomes (seen anywhere, even in denominator) (A1)

valid approach of listing ways to get sum of 5, showing at least two pairs (M1)

e.g. (1, 4)(2, 3), (1, 4)(4, 1), (1, 4)(4, 1), (2, 3)(3, 2) , lattice diagram

A1 N3

[3 marks]

Examiners reportWhile many candidates were successful at part (a), far fewer recognized the binomial distribution in the second part of the problem.

45b. [2 marks]Jan tosses the two dice 8 times. Find the probability that she wins 3 prizes.

Markschemerecognizing binomial probability (M1)

e.g. , binomial pdf,

A1 N2

[2 marks]

Examiners reportWhile many candidates were successful at part (a), far fewer recognized the binomial distribution in the second part of the problem.

Those who did not obtain the correct answer at part (a) often scored partial credit by either drawing a table to represent the samplespace or by noting relevant pairs.

46a. [2 marks]

Let X be normally distributed with mean 100 cm and standard deviation 5 cm.

On the diagram below, shade the region representing .

Markscheme

A1A1 N2

Note: Award A1 for vertical line to right of mean, A1 for shading to right of their vertical line.

Examiners reportMost candidates did very well on part (a), shading the area under the normal curve.

46b. [2 marks]Given that , find the value of .

Markschemeevidence of recognizing symmetry (M1)

e.g. is one standard deviation above the mean so is one standard deviation below the mean, shading the corresponding part,

A1 N2

[2 marks]

Examiners reportNot all candidates realised that the problem could be solved by only using the symmetry of the normal distribution curve and theinformation given. Some of them saw the need to use tables and others just left it blank.

Candidates were only moderately successful on parts (b) and (c), which required understanding of the symmetry of the curve. Manycandidates resorted to formulae or tables instead of reasoning through the question.

46c. [2 marks]Given that (correct to two significant figures), find .

Markschemeevidence of using complement (M1)

e.g. ,

A1 N2

[2 marks]

Examiners reportNot all candidates realised that the problem could be solved by only using the symmetry of the normal distribution curve and theinformation given. Some of them saw the need to use tables and others just left it blank.

Candidates were only moderately successful on parts (b) and (c), which required understanding of the symmetry of the curve. Manycandidates resorted to formulae or tables instead of reasoning through the question.

47a. [4 marks]

In a class of 100 boys, 55 boys play football and 75 boys play rugby. Each boy must play at least one sport from football and rugby.

One boy is selected at random.

(i) Find the probability that he plays only one sport.

(ii) Given that the boy selected plays only one sport, find the probability that he plays rugby.

Markscheme(i) METHOD 1

evidence of using complement, Venn diagram (M1)

e.g. ,

A1 N2

METHOD 2

attempt to find P(only one sport) , Venn diagram (M1)

e.g.

A1 N2

(ii) A2 N2

[4 marks]

Examiners reportOverall, this question was very well done. There were some problems with the calculation of conditional probability, where aconsiderable amount of candidates tried to use a formula instead of using its concept and analysing the problem. It is the kind ofquestion where it can be seen if the concept is not clear to candidates.

47b. [2 marks]Let A be the event that a boy plays football and B be the event that a boy plays rugby.

Explain why A and B are not mutually exclusive.

Markschemevalid reason in words or symbols (R1)

e.g. if mutually exclusive, if not mutually exclusive

correct statement in words or symbols A1 N2

e.g. , , , some students play both sports, sets intersect

[2 marks]

Examiners reportIn part (c), candidates were generally able to explain in words why events were mutually exclusive, though many gave the wrongvalues for P(A) and P(B).

47c. [3 marks]Show that A and B are not independent.

Markschemevalid reason for independence (R1)

e.g. ,

correct substitution A1A1 N3

e.g. ,

[3 marks]

Examiners reportThere was a great amount of confusion between the concepts of independent and mutually exclusive events. In part (d), theexplanations often referred to mutually exclusive events.

It was evident that candidates need more practice with questions like (c) and (d).

Some students equated probabilities and number of elements, giving probabilities greater than 1.

48a. [1 mark]

A multiple choice test consists of ten questions. Each question has five answers. Only one of the answers is correct. For each question, Joserandomly chooses one of the five answers.

Find the expected number of questions Jose answers correctly.

Markscheme A1 N1

[1 mark]

Examiners reportMost candidates were able to find the mean by applying various methods. Although many recognised binomial probability, fewerwere able to use the GDC effectively.

48b. [2 marks]Find the probability that Jose answers exactly three questions correctly.

Markschemeevidence of appropriate approach involving binomial (M1)

e.g. , ,

A1 N2

[2 marks]

Examiners reportMost candidates were able to find the mean by applying various methods. Although many recognised binomial probability, fewerwere able to use the GDC effectively.

48c. [3 marks]Find the probability that Jose answers more than three questions correctly.

MarkschemeMETHOD 1

(A1)

evidence of using the complement (seen anywhere) (M1)

e.g. any probability ,

A1 N2

METHOD 2

recognizing that (M1)

e.g. summing probabilities from to

correct expression or values (A1)

e.g.

A1 N2

[3 marks]

Examiners reportPart (c) was problematic in some cases but most candidates recognized that either a sum of probabilities or the complement wasrequired. Many misinterpreted "more than three" as inclusive of three, and so obtained incorrect answers. When adding individualprobabilities, some candidates used three or fewer significant figures, which resulted in an incorrect final answer due to prematurerounding.

49. [7 marks]Consider the independent events A and B . Given that , and , find .

MarkschemeMETHOD 1

for independence (R1)

expression for , indicating (A1)

e.g. ,

substituting into (M1)


e.g. ,

correct solutions to the equation (A2)

e.g. , (accept the single answer )

A1 N6

[7 marks]

METHOD 2

for independence (R1)

expression for , indicating (A1)

e.g. ,

substituting into (M1)


e.g. ,

correct solutions to the equation (A2)

e.g. 0.4, 2.6 (accept the single answer 0.4)

(accept if x set up as ) A1 N6

[7 marks]

Examiners reportMany candidates confused the concept of independence of events with mutual exclusivity, mistakenly trying to use the formula

. Those who did recognize that were often able to find the correct equation, butmany were unable to use their GDC to solve it. A few provided two answers without discarding the value greater than one.

50a. [1 mark]

The letters of the word PROBABILITY are written on 11 cards as shown below.

Two cards are drawn at random without replacement.

Let A be the event the first card drawn is the letter A.

Let B be the event the second card drawn is the letter B.

Find .

Markscheme A1 N1

[1 mark]

Examiners reportMost candidates answered part (a) correctly.


Markscheme A2 N2

[2 marks]

Examiners reportFew candidates used the concept of "B given A" to simply "write down" the answer of . Instead, most reached for the formula inthe booklet, with which few were successful.


Markschemerecognising that (M1)

correct values (A1)

e.g.

A1 N3

[3 marks]

Examiners reportFew also made the connection that part (c) could be answered using both previous answers. Many found correctly evenwhen answering part (b) incorrectly, although some candidates did not decrease the denominator for the second event.

51a. [2 marks]

A random variable X is distributed normally with mean 450 and standard deviation 20.

Find .

Markschemeevidence of attempt to find (M1)

e.g.

A1 N2

[2 marks]

Examiners reportIt remains very clear that some centres still do not give appropriate attention to the normal distribution. This is a major cause forconcern. Most candidates had been taught the topic but many had difficulty understanding the difference between , , and .Very little working was shown which demonstrated understanding. Although the GDC was used extensively, candidates oftenworked with the wrong tail and did not write their answers correct to 3 significant figures.

51b. [4 marks]Given that , find .

Markschemeevidence of using the complement (M1)

e.g. 0.73,

(A1)

setting up equation (M1)

e.g.

A1 N3

[4 marks]

Examiners reportIt remains very clear that some centres still do not give appropriate attention to the normal distribution. This is a major cause forconcern. Most candidates had been taught the topic but many had difficulty understanding the difference between , , and .Very little working was shown which demonstrated understanding. Although the GDC was used extensively, candidates oftenworked with the wrong tail and did not write their answers correct to 3 significant figures.

Many candidates had trouble with part (b), a majority never found the complement, instead using their GDCs to calculate the result,which many times was finding a for instead of for . Many others substituted the values of or

into the equation, instead of the -scores.

52a. [4 marks]

In any given season, a soccer team plays 65 % of their games at home.

When the team plays at home, they win 83 % of their games.

When they play away from home, they win 26 % of their games.

The team plays one game.

Find the probability that the team wins the game.

Markschemeappropriate approach (M1)

e.g. tree diagram or a table

(M1)

A1

(or 0.631) A1 N2

[4 marks]

Examiners reportPart (a) was nearly always correctly answered by those who attempted the question, but part (b) (conditional probability) was poorlydone. A surprisingly small number of students drew a tree diagram in part (a) and those who did answered this part and part (b) well.Many found the correct complement in part (b) but could not make any further progress.

52b. [4 marks]If the team does not win the game, find the probability that the game was played at home.

Markschemeevidence of using complement (M1)

e.g. , 0.3695

choosing a formula for conditional probability (M1)

e.g.

correct substitution

e.g. A1

P(home) = 0.299 A1 N3

[4 marks]

Examiners reportPart (b) (conditional probability) was poorly done. A surprisingly small number of students drew a tree diagram in part (a) and thosewho did answered this part and part (b) well. Many found the correct complement in part (b) but could not make any further progress.

53a. [3 marks]

A fisherman catches 200 fish to sell. He measures the lengths, l cm of these fish, and the results are shown in the frequency table below.

Calculate an estimate for the standard deviation of the lengths of the fish.

Markschemeevidence of using mid-interval values (5, 15, 25, 35, 50, 67.5, 87.5) (M1)

(cm) A2 N3

[3 marks]

Examiners reportPart (a) defeated the vast majority of candidates who clearly had not been taught data entry. Some schools had attempted to teach howto use a formula rather than the GDC to find the standard deviation and their students invariably used this formula incorrectly. Use ofthe GDC was not only expected but should be emphasized as stated in the syllabus.

53b. [6 marks]A cumulative frequency diagram is given below for the lengths of the fish.

Use the graph to answer the following.

(i) Estimate the interquartile range.

(ii) Given that of the fish have a length more than , find the value of k.

Markscheme(i) , (A1)(A1)

(accept any notation that suggests the interval 15 to 40) A1 N3

(ii) METHOD 1

have a length less than k (A1)

(A1)

(cm) A1 N2

METHOD 2

(A1)

(A1)

(cm) A1 N2

[6 marks]

Examiners reportPart (b) revealed poor understanding of cumulative frequency and the IQR was often reported as an interval.

53c. [2 marks]In order to sell the fish, the fisherman classifies them as small, medium or large.

Small fish have a length less than .

Medium fish have a length greater than or equal to but less than .

Large fish have a length greater than or equal to .

Write down the probability that a fish is small.

Markscheme (M1)

A1 N2

[2 marks]

Examiners reportThis was generally answered well although a number of candidates had difficulty with using the formula for expected value.

53d. [2 marks]The cost of a small fish is , a medium fish , and a large fish .

Copy and complete the following table, which gives a probability distribution for the cost .

Markscheme

A1A1 N2

[2 marks]


53e. [2 marks]Find .

Markschemecorrect substitution (of their p values) into formula for (A1)

e.g.

(accept ) A1 N2

[2 marks]


54a. [2 marks]

Two boxes contain numbered cards as shown below.

Two cards are drawn at random, one from each box.

Copy and complete the table below to show all nine equally likely outcomes.

Markscheme

A2 N2

[2 marks]

Examiners reportMost candidates completed parts (a), (b) and (c) successfully.

54b. [2 marks]Let S be the sum of the numbers on the two cards.

Find the probability of each value of S.

Markscheme , , , A2 N2

[2 marks]

Examiners reportMost candidates completed part (b) successfully.

54c. [3 marks]Find the expected value of S.

Examiners reportMany found the expected value correctly, while some showed difficulty with the arithmetic.

54d. [3 marks]Anna plays a game where she wins if S is even and loses if S is odd.

Anna plays the game 36 times. Find the amount she expects to have at the end of the 36 games.

MarkschemeMETHOD 1

correct expression for expected gain E(A) for 1 game (A1)

e.g.

amount at end = expected gain for 1 game (M1)

= 200 (dollars) A1 N2

METHOD 2

attempt to find expected number of wins and losses (M1)

e.g. ,

attempt to find expected gain E(G) (M1)

e.g.

(dollars) A1 N2

[3 marks]

Examiners reportThis was often left blank or only superficially attempted. Some found the expected value but did not answer the question about theamount of money.

55a. [1 mark]

The following diagram is a box and whisker plot for a set of data.

The interquartile range is 20 and the range is 40.

Write down the median value.

Markscheme18 A1 N1

[1 mark]

Examiners reportMost candidates were able to find the values for the median, lower quartile, and point b. A large majority answered this questioncorrectly.

55b. [4 marks]Find the value of

(i) ;

(ii) .

Markscheme(i) 10 A2 N2

(ii) 44 A2 N2

[4 marks]

Examiners reportMost candidates were able to find the values for the median, lower quartile, and point b. A large majority answered this questioncorrectly.

56a. [2 marks]

A van can take either Route A or Route B for a particular journey.

If Route A is taken, the journey time may be assumed to be normally distributed with mean 46 minutes and a standard deviation 10 minutes.

If Route B is taken, the journey time may be assumed to be normally distributed with mean minutes and standard deviation 12 minutes.

For Route A, find the probability that the journey takes more than minutes.

Markscheme

A2 N2

[2 marks]

Examiners reportA significant number of students clearly understood what was asked in part (a) and used the GDC to find the result.

56b. [3 marks]For Route B, the probability that the journey takes less than minutes is .

Find the value of .

Markschemecorrect approach (A1)

e.g. , sketch

(A1)

A1 N2

[3 marks]

Examiners reportIn part (b), many candidates set the standardized formula equal to the probability ( ), instead of using the corresponding z-score.Other candidates used the solver on their GDC with the inverse norm function.

56c. [3 marks]The van sets out at 06:00 and needs to arrive before 07:00.

(i) Which route should it take?

(ii) Justify your answer.

Markscheme(i) route A A1 N1

(ii) METHOD 1

A1

valid reason R1

e.g. probability of A getting there on time is greater than probability of B

N2

METHOD 2

A1

valid reason R1

e.g. probability of A getting there late is less than probability of B

N2

[3 marks]

Examiners reportA common incorrect approach in part (c) was to attempt to use the means and standard deviations for justification, although manycandidates successfully considered probabilities.

56d. [5 marks]On five consecutive days the van sets out at 06:00 and takes Route B. Find the probability that

(i) it arrives before 07:00 on all five days;

(ii) it arrives before 07:00 on at least three days.

Markscheme(i) let X be the number of days when the van arrives before 07:00

(A1)

A1 N2

(ii) METHOD 1

evidence of adding correct probabilities (M1)

e.g.

correct values (A1)

A1 N3

METHOD 2

evidence of using the complement (M1)

e.g. ,

correct values (A1)

A1 N3

[5 marks]

Examiners reportA pleasing number of candidates recognized the binomial probability and made progress on part (d).

57a. [1 mark]

Let A and B be independent events, where and .

Write down an expression for .

Markscheme A1 N1

[1 mark]

Examiners reportThis question was well done by most candidates.

57b. [4 marks]Given that ,

(i) find x ;

(ii) find .

Markscheme(i) evidence of using (M1)


e.g. ,

A1 N2

(ii) A1 N1

[4 marks]

Examiners reportThis question was well done by most candidates. When errors were made, candidates confused the terms "independent" and"mutually exclusive" and did not subtract the intersection when finding .

57c. [1 mark]Hence, explain why A and B are not mutually exclusive.

Markschemevalid reason, with reference to R1 N1

e.g.

[1 mark]

Examiners reportCandidates should also be aware of the command term "hence" used in part (c) where they were expected to provide a reason thatinvolved from their work in part (b). It seemed that many turned to the formula in the booklet instead of considering theconceptual meaning of the term.

58a. [1 mark]

The following is a cumulative frequency diagram for the time t, in minutes, taken by 80 students to complete a task.

Write down the median.

Markschememedian A1 N1

[1 mark]

Examiners reportThis question was answered successfully by a majority of candidates. A common error was to use values of 20 and 60 for the lowerand upper quartiles. Some were careless when reading the graph scale and wrote incorrect answers as a result.

58b. [3 marks]Find the interquartile range.

Markschemelower quartile , upper quartile (A1)(A1)

A1 N3

[3 marks]


58c. [2 marks]Complete the frequency table below.

Markscheme

A1A1 N2

[2 marks]


59a. [2 marks]

The probability of obtaining heads on a biased coin is 0.18. The coin is tossed seven times.

Find the probability of obtaining exactly two heads.

Markschemeevidence of using binomial probability (M1)

e.g.

A1 N2

[2 marks]

Examiners reportCandidates who recognized binomial probability answered this question very well, using their GDC to perform the final calculations.

59b. [3 marks]Find the probability of obtaining at least two heads.

MarkschemeMETHOD 1

evidence of using the complement M1

e.g.

(A1)

A1 N2

METHOD 2

evidence of attempting to sum probabilities M1

e.g. ,

correct values for each probability (A1)

e.g.

A1 N2

[3 marks]

Examiners reportSome candidates misinterpreted the meaning of "at least two" in part (b), and instead found . Others wrote down a correctinterpretation but accumulated to in their GDC (e.g. binomcdf (7, 0.18, 2)). Still, the number of candidates who either left thisquestion blank or approached the question without binomial considerations suggests that this topic continues to be neglected in somecentres.

60a. [3 marks]

The scores of a test given to students are normally distributed with a mean of 21. of the students have scores less than 23.7.

Find the standard deviation of the scores.

Markschemeevidence of approach (M1)

e.g. finding , using


e.g. , graph A1

[3 marks]

Examiners reportCandidates who clearly understood the nature of normal probability answered this question cleanly. A common misunderstandingwas to use the value of 0.8 as a z-score when finding the standard deviation.

60b. [4 marks]A student is chosen at random. This student has the same probability of having a score less than 25.4 as having a score greaterthan b.

(i) Find the probability the student has a score less than 25.4.

(ii) Find the value of b.

Markscheme(i) evidence of attempting to find (M1)

e.g. using

A1 N2

(ii) evidence of recognizing symmetry (M1)

e.g. , using A1 N2

[4 marks]

Examiners reportMany correctly used their GDC to find the probability in part (b). Fewer used some aspect of the symmetry of the curve to find avalue for b.

61a. [3 marks]

In a school with 125 girls, each student is tested to see how many sit-up exercises (sit-ups) she can do in one minute. The results are given inthe table below.

(i) Write down the value of p.

(ii) Find the value of q.

Markscheme(i) A1 N1

(ii) for evidence of using sum is 125 (or ) (M1)

A1 N2

[3 marks]

Examiners reportPart (a) of this question was well done.

61b. [2 marks]Find the median number of sit-ups.

Markschemeevidence of median position (M1)

e.g. 63rd student,

median is 17 (sit-ups) A1 N2

[2 marks]

Examiners reportFinding the median seemed to be the most difficult for the candidates. Most had the idea that it was in the middle but did not knowhow to find the value.

61c. [2 marks]Find the mean number of sit-ups.

Markschemeevidence of substituting into (M1)

e.g. ,

mean A1 N2

[2 marks]

Examiners reportWhen calculating the mean, many ignored the frequencies.

62a. [2 marks]

A factory makes switches. The probability that a switch is defective is 0.04. The factory tests a random sample of 100 switches.

Find the mean number of defective switches in the sample.

Markschemeevidence of binomial distribution (may be seen in parts (b) or (c)) (M1)

e.g. np,

A1 N2

[2 marks]

Examiners reportPart (a) was handled well by most students.

62b. [2 marks]Find the probability that there are exactly six defective switches in the sample.

Markscheme (A1)

A1 N2

[2 marks]

Examiners reportAlthough this question was a rather straightforward question on binomial distribution, parts (b) and(c) seemed to cause muchdifficulty.

62c. [3 marks]Find the probability that there is at least one defective switch in the sample.

Markschemefor evidence of appropriate approach (M1)

e.g. complement,

(A1)

A1 N2

[3 marks]

Examiners reportAlthough this question was a rather straightforward question on binomial distribution, parts (b) and(c) seemed to cause muchdifficulty. In part (c), finding at least one defective switch, many forgot to take the complement.

63a. [4 marks]

A box contains a large number of biscuits. The weights of biscuits are normally distributed with mean and standard deviation .

One biscuit is chosen at random from the box. Find the probability that this biscuit

(i) weighs less than ;

(ii) weighs between and .

Markscheme

(i) (M1)

A1 N2


e.g. symmetry,

(tables 0.955) A1 N2

Note: Award M1A1(AP) if candidates refer to 2 standard deviations from the mean, leading to 0.95.

[4 marks]

Examiners reportThose that understood the normal distribution did well on parts (a) and (bi).

63b. [5 marks]Five percent of the biscuits in the box weigh less than d grams.

(i) Copy and complete the following normal distribution diagram, to represent this information, by indicating d, and shading theappropriate region.

(ii) Find the value of d.

Markscheme(i)

A1A1 N2

Note: Award A1 for d to the left of the mean, A1 for area to the left of d shaded.

(ii) (A1)

(M1)

A1 N3

[5 marks]

Examiners reportThose that understood the normal distribution did well on parts (a) and (bi). Parts (bii) and (c) proved to be a little more difficult. Inparticular, in part (bii) the z-score was incorrectly set equal to 0.05 and in part (c), 0.2 was used instead of the z-score. For those whohad a good grasp of the concept of normal distributions the entire question was quite accessible and full marks were gained.

63c. [4 marks]The weights of biscuits in another box are normally distributed with mean and standard deviation . It is known that of the biscuits in this second box weight less than .

Find the value of .

Markscheme

(M1)

A1

(M1)

A1 N3

[4 marks]

Examiners reportThose that understood the normal distribution did well on parts (a) and (bi). Parts (bii) and (c) proved to be a little more difficult. Inparticular, in part (bii) the z-score was incorrectly set equal to 0.05 and in part (c), 0.2 was used instead of the z-score. For those whohad a good grasp of the concept of normal distributions the entire question was quite accessible and full marks were gained.

64a. [2 marks]

A box contains 100 cards. Each card has a number between one and six written on it. The following table shows the frequencies for eachnumber.

Calculate the value of k.

Markschemeevidence of using (M1)

A1 N2

[2 marks]

Examiners reportFrequencies and median seemed well understood, but quartiles and inter-quartile range less so.

64b. [5 marks]Find

(i) the median;

(ii) the interquartile range.

Markscheme(i) evidence of median position (M1)

e.g. 50th item,

A1 N2

(ii) and (A1)(A1)

(accept 1 to 5 or , etc.) A1 N3

[5 marks]

Examiners reportFrequencies and median seemed well-understood, but quartiles and interquartile range less so. A few students, probably based on pastpapers, drew cumulative frequency diagrams, generating slightly different answers for median and quartiles.

65a. [4 marks]

There are 20 students in a classroom. Each student plays only one sport. The table below gives their sport and gender.

One student is selected at random.

(i) Calculate the probability that the student is a male or is a tennis player.

(ii) Given that the student selected is female, calculate the probability that the student does not play football.

Markscheme(i) correct calculation (A1)

e.g. ,

A1 N2

(ii) correct calculation (A1)

e.g. ,

A1 N2

[4 marks]

Examiners reportMany candidates had difficulty with this question, usually as a result of seeking to solve the problem by formula instead of lookingcarefully at the table frequencies.

65b. [3 marks]Two students are selected at random. Calculate the probability that neither student plays football.

Markscheme , A1

A1

A1 N1

[3 marks]

Examiners reportA very common error in part (b) was to assume identical probabilities for each selection instead of dependent probabilities wherethere is no replacement.

66a. [2 marks]

A four-sided die has three blue faces and one red face. The die is rolled.

Let B be the event a blue face lands down, and R be the event a red face lands down.

Write down

(i) P(B);(ii) P(R).

Markscheme(i) P(B) A1 N1

(ii) P(R) A1 N1

[2 marks]

Examiners reportThis was the most difficult of the extended response questions for the candidates. Finding s and t correctly in part (b) was difficult,with many confused between writing appropriate probabilities on a single branch compared to at the final end of a multiple branch.Many candidates had no idea what to write for a probability distribution and those who did often had probabilities that did not sum to1. Candidates who wrote a probability distribution often could correctly compute the expected value. The final part was the mostchallenging, but some good answers were seen. The most common error was not recognizing that there were two different ways ofwinning.

66b. [2 marks]If the blue face lands down, the die is not rolled again. If the red face lands down, the die is rolled once again. This isrepresented by the following tree diagram, where p, s, t are probabilities.

Find the value of p, of s and of t.

Markscheme A1 N1

, A1 N1

[2 marks]


66c. [3 marks]Guiseppi plays a game where he rolls the die. If a blue face lands down, he scores 2 and is finished. If the red face landsdown, he scores 1 and rolls one more time. Let X be the total score obtained.

(i) Show that .(ii) Find .

Markscheme(i)

A1

AG N0

(ii) (A1)

A1 N2

[3 marks]


66d. [5 marks](i) Construct a probability distribution table for X.

(ii) Calculate the expected value of X.

Markscheme(i)

A2 N2

(ii) evidence of using (M1) (A1)

A1 N2

[5 marks]


66e. [4 marks]If the total score is 3, Guiseppi wins . If the total score is 2, Guiseppi gets nothing.

Guiseppi plays the game twice. Find the probability that he wins exactly .

Markschemewin scores 3 one time, 2 other time (M1)

(seen anywhere) A1

evidence of recognising there are different ways of winning (M1)e.g. , ,

A1 N3

[4 marks]


67a. [4 marks]

The following table shows the probability distribution of a discrete random variable X.

Find the value of k.



e.g. ,

A2 N2

[4 marks]

Examiners reportA good number of candidates answered this question well, although some incorrectly set the sum of the probabilities to zero insteadof one, suggesting rote recognition of a quadratic equal to zero. Many candidates recognized that only the positive value for k wasappropriate and correctly indicated this in their working. Many went on to find the correct expected value as well, although at timescandidates wrote the formula from the information booklet without making use of it, thus earning no marks.

67b. [3 marks]Find the expected value of X.



e.g.

A1 N2

[3 marks]

Examiners reportA good number of candidates answered this question well, although some incorrectly set the sum of the probabilities to zero insteadof one, suggesting rote recognition of a quadratic equal to zero. Many candidates recognized that only the positive value for k wasappropriate and correctly indicated this in their working. Many went on to find the correct expected value as well, although at timescandidates wrote the formula from the information booklet without making use of it, thus earning no marks.

68a. [2 marks]

The heights of certain plants are normally distributed. The plants are classified into three categories.

The shortest are in category A.

The tallest are in category C.

All the other plants are in category B with heights between and .

Complete the following diagram to represent this information.

Markscheme

A1A1 N2

Notes: Award A1 for three regions (may be shown by lines or shading), A1 for clear labelling of two regions (may be shown bypercentages or categories). r and t need not be labelled, but if they are, they may be interchanged.

[2 marks]

Examiners reportMany candidates shaded or otherwise correctly labelled the appropriate regions in the normal curve.

68b. [5 marks]Given that the mean height is and the standard deviation , find the value of r and of t.

MarkschemeMETHOD 1

(A1)

A1 N2

(= 0.8962) (may be seen later) A1

(A1)

A1 N2

METHOD 2

finding z-values A1A1

evidence of setting up one standardised equation (M1)

e.g. ,

, A1A1 N2N2

[5 marks]

Examiners reportAlthough many candidates shaded or otherwise correctly labelled the appropriate regions in the normal curve, far fewer could applytechniques of normal probabilities to achieve correct results in part (b). Many set the standardized formula equal to the probabilitiesinstead of the appropriate z-scores, which can be found either by the use of tables or the GDC. Others simply left this part blank,which suggests a lack of preparation for such “inverse” types of questions in a normal distribution.

69a. [2 marks]

Paula goes to work three days a week. On any day, the probability that she goes on a red bus is .

Write down the expected number of times that Paula goes to work on a red bus in one week.

Markschemeevidence of binomial distribution (seen anywhere) (M1)

e.g.

( ) A1 N2

[2 marks]

Examiners reportMany candidates did not recognize the binomial nature of this question, suggesting an overall lack of preparation with this topic.Many used 7 days instead of 3 but could still earn marks in follow-through if working was shown. Those who could use their GDCeffectively often answered correctly.

69b. [2 marks]In one week, find the probability that she goes to work on a red bus on exactly two days.

Markscheme (A1)

A1 N2

[2 marks]

Examiners reportMany candidates did not recognize the binomial nature of this question, suggesting an overall lack of preparation with this topic.Many used 7 days instead of 3 but could still earn marks in follow-through if working was shown. Those who could use their GDCeffectively often answered correctly.

69c. [3 marks]In one week, find the probability that she goes to work on a red bus on at least one day.

Markschemeevidence of appropriate approach M1

e.g. complement, , adding probabilities

(A1)

A1 N2

[3 marks]

Examiners reportMany candidates did not recognize the binomial nature of this question, suggesting an overall lack of preparation with this topic.Many used 7 days instead of 3 but could still earn marks in follow-through if working was shown. Those who could use their GDCeffectively often answered correctly, although in part (c) some candidates misinterpreted the meaning of “at least one” and foundeither or .

70a. [2 marks]

There are nine books on a shelf. For each book, x is the number of pages, and y is the selling price in pounds (£). Let r be the correlationcoefficient.

Write down the possible minimum and maximum values of r .

Markschememin value of r is , max value of r is 1 A1A1 N2

[2 marks]

Examiners report[N/A]

70b. [1 mark]Given that , which of the following diagrams best represents the data.

MarkschemeC A1 N1

[1 mark]


70c. [2 marks]

For the data in diagram D , which two of the following expressions describe the correlation between x and y ?

perfect, zero, linear, strong positive, strong negative, weak positive, weak negative

Markschemelinear, strong negative A1A1 N2

[2 marks]


71a. [2 marks]

A data set has a mean of 20 and a standard deviation of 6.

Each value in the data set has 10 added to it. Write down the value of

(i) the new mean;

(ii) the new standard deviation.

Markscheme(i) new mean is A1 N1

(ii) new sd is 6 A1 N1

[2 marks]


71b. [3 marks]Each value in the original data set is multiplied by 10.

(i) Write down the value of the new mean.

(ii) Find the value of the new variance.

Markscheme(i) new mean is A1 N1

(ii) METHOD 1

variance is 36 A1

new variance is A1 N2

METHOD 2

new sd is 60 A1

new variance is A1 N2

[3 marks]


72a. [6 marks]

Two standard six-sided dice are tossed. A diagram representing the sample space is shown below.

Let be the sum of the scores on the two dice.

(i) Find .

(ii) Find .

(iii) Find .

Markscheme(i) number of ways of getting is 5 A1

A1 N2

(ii) number of ways of getting is 21 A1

A1 N2

(iii) A2 N2

[6 marks]


72b. [8 marks]Elena plays a game where she tosses two dice.

If the sum is 6, she wins 3 points.

If the sum is greater than 6, she wins 1 point.

If the sum is less than 6, she loses k points.

Find the value of k for which the game is fair.

Markschemeattempt to find M1

e.g.

A1

fair game if (may be seen anywhere) R1

attempt to substitute into formula M1

e.g.

correct substitution into A1

e.g.

work towards solving M1

e.g.

A1

A1 N4

[8 marks]


73a. [3 marks]

A random variable X is distributed normally with mean 450. It is known that .

Represent all this information on the following diagram.

Markscheme

A1A1A1 N3

Note: Award A1 for 450 , A1 for a to the right of the mean, A1 for area 0.27 .

[3 marks]


73b. [3 marks]Given that the standard deviation is 20, find a . Give your answer correct to the nearest whole number.

Markschemevalid approach M1

e.g. , 0.73

A1

A1 N3

[3 marks]


74a. [4 marks]

The probability of obtaining heads on a biased coin is 0.4. The coin is tossed 600 times.

(i) Write down the mean number of heads.

(ii) Find the standard deviation of the number of heads.

Markscheme(i) recognizing binomial with , M1

A1 N2

(ii) correct substitution into formula for variance or standard deviation A1

e.g. 144,

sd = 12 A1 N1

[4 marks]


74b. [3 marks]Find the probability that the number of heads obtained is less than one standard deviation away from the mean.

Markschemeattempt to find range of values M1

e.g.

evidence of correct approach A1

e.g.

A1 N2

[3 marks]


75a. [2 marks]

Each day, a factory recorded the number ( ) of boxes it produces and the total production cost ( ) dollars. The results for nine days areshown in the following table.

Write down the equation of the regression line of y on x .

Markscheme A1A1 N2

[2 marks]


75b. [2 marks]Use your regression line from part (a) as a model to answer the following.

Interpret the meaning of

(i) the gradient;

(ii) the y-intercept.

Markscheme(i) additional cost per box (unit cost) A1 N1

(ii) fixed costs A1 N1

[2 marks]


75c. [2 marks]Estimate the cost of producing 60 boxes.

Printed for sek-catalunya

© International Baccalaureate Organization 2014 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markschemeattempt to substitute into regression equation M1

e.g. ,

(accept from 3 s.f. values) A1 N2

[2 marks]


75d. [3 marks]The factory sells the boxes for $19.99 each. Find the least number of boxes that the factory should produce in one day in orderto make a profit.

Markschemesetting up inequality (accept equation) M1

e.g.

A1

13 boxes (accept 14 from , using 3 s.f. values) A1 N2

Note: Exception to the FT rule: if working shown, award the final A1 for a correct integer solution for their value of x.

[3 marks]


75e. [4 marks]Comment on the appropriateness of using your model to

(i) estimate the cost of producing 5000 boxes;

(ii) estimate the number of boxes produced when the total production cost is $540.

Markscheme(i) this would be extrapolation, not appropriate R1R1 N2

(ii) this regression line cannot predict x from y, not appropriate R1R1 N2

[4 marks]


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